Question

Derive 1st, 2nd and 3rd equation of motion in details .. NO SPAM

Answer 1

here is ur answer achu .....

first equation of motion :  v =  u + at

acceleration = final velocity - initial velocity /  2

so ,

at = v - u (  this is because we take ` t `  with ` a ` so it came at .

then we take v to the next side and we get the first equation as :

v = u + at

second equation of motion :  s = ut + 1/2 at square

in a graph area of OABC = area of rectangle OADC and area of Δ ABC

s = OA * OC + 1/2 * AD* BD

OA = U

OC = t

AD = t

BD = v-u

S = u * t + 1/2 * t ( v-u )

=  u * t + 1/2 *t *at

S = ut + 1/2 at square

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Answer 2

The three kinematic equations are possible only for the following conditions:

• The acceleration must be a constant and does not vary with time.
• The body in motion has an initial velocity u at time 0 seconds and a final velocity v at time t seconds.

Since the acceleration is a constant, the velocity - time graph of the body will be as shown below.

$\setlength{\unitlength}{1cm}\begin {picture}(5,5)\put(0,0){\vector (1,0){5}}\put(0,0){\vector (0,1){5}}\put(5.2,-0.2){ t }\put(-0.3,5.2){ v }\put (0,1){\vector (1,1){3.5}}\put(-0.85,1){ (0,u) }\multiput(2,3)(0,-0.4){8}{\line (0,-1){0.2}}\put(1.3,3.2){ (t,v) }\put(1.6,-0.45){ (t,0) }\put(-0.85,-0.4){ (0,0) }\end {picture}$

Here the slope of the graph gives acceleration. So,

$a=\dfrac {v-u}{t-0}\\\\\\v-u=at\\\\\\\boxed {v=u+at}$

Thus first kinematic equation is derived.

We know that displacement is the integral of the velocity wrt time. So,

$\displaystyle s=\int\limits_{0}^{t}v\ dt\\\\\\s=\int\limits_{0}^{t}(u+at)\ dt\\\\\\s=\int\limits_{0}^{t}u\ dt+\int\limits_{0}^{t}at\ dt\\\\\\s=u\left [t\right]_0^t+a\left [\dfrac {t^2}{2}\right]_0^t\\\\\\\boxed {s=ut+\dfrac {1}{2}at^2}$

Thus second kinematic equation is also derived.

The third kinematic equation can be derived in two ways.

First, recall the first kinematic equation.

$v=u+at\\\\\\v^2=(u+at)^2\\\\\\v^2=u^2+2uat+a^2t^2\\\\\\v^2=u^2+2a\left (ut+\dfrac {1}{2}at^2\right)\\\\\\\boxed{v^2=u^2+2as}$

Or, second, the third kinematic equation can be derived by the displacement of the body, which is given by the area under the graph.

Here the region between the graph and time axis is a trapezium. So the displacement will be,

$s=\dfrac {1}{2}t(v+u)\\\\\\s=\dfrac {(v+u)t}{2}\quad\longrightarrow\quad (1)$

But,

$v=u+at\quad\implies\quad t=\dfrac {v-u}{a}$

Then (1) becomes,

$s=\dfrac {(v+u)(v-u)}{2a}\\\\\\s=\dfrac {v^2-u^2}{2a}\\\\\\v^2-u^2=2as\\\\\\\boxed {v^2=u^2+2as}$

Thus third kinematic equation is also derived.

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